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Let me call $L$ your Lie algebra $\mathbb{R}^n$. The condition for a subspace $V\subset L$ to be a subalgebra is a closed condition: you want the map $V\otimes V\rightarrow L/V$ deduced from the bracket to be zero. In other words, let $\mathcal{V}$ be the universal $k$-plane bundle on $Gr(k,n)$, and $\mathcal{Q}$ the universal quotient; $Gr(k,n)_{L}$ is the zero locus of the map $\mathcal{V}\otimes \mathcal{V}\rightarrow \mathcal{Q} $ induced by the bracket. Thus $Gr(k,n)_{L}$ is a closed subset of $Gr(k,n)$, hence it is always compact.